Question: $\dfrac{ 4t - 3u }{ 8 } = \dfrac{ -7t + 8v }{ 9 }$ Solve for $t$.
Explanation: Multiply both sides by the left denominator. $\dfrac{ 4t - 3u }{ {8} } = \dfrac{ -7t + 8v }{ 9 }$ ${8} \cdot \dfrac{ 4t - 3u }{ {8} } = {8} \cdot \dfrac{ -7t + 8v }{ 9 }$ $4t - 3u = {8} \cdot \dfrac { -7t + 8v }{ 9 }$ Multiply both sides by the right denominator. $4t - 3u = 8 \cdot \dfrac{ -7t + 8v }{ {9} }$ ${9} \cdot \left( 4t - 3u \right) = {9} \cdot 8 \cdot \dfrac{ -7t + 8v }{ {9} }$ ${9} \cdot \left( 4t - 3u \right) = 8 \cdot \left( -7t + 8v \right)$ Distribute both sides ${9} \cdot \left( 4t - 3u \right) = {8} \cdot \left( -7t + 8v \right)$ ${36}t - {27}u = -{56}t + {64}v$ Combine $t$ terms on the left. ${36t} - 27u = -{56t} + 64v$ ${92t} - 27u = 64v$ Move the $u$ term to the right. $92t - {27u} = 64v$ $92t = 64v + {27u}$ Isolate $t$ by dividing both sides by its coefficient. ${92}t = 64v + 27u$ $t = \dfrac{ 64v + 27u }{ {92} }$